An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size
Answers
Answer:
Given: u=−20 cm
f=R/2=15 cm
Size of image, h=5 cm
Let size of image be h
′
From mirror formula,
u1 +v1=−20
1 + v1 = 15
1 v= 760
=8.57 cm
The image is formed 8.57cm behind the mirror. It is virtual and erect.
Magnification, m=−v/u=8.57/20=0.428
Also, m=h
′
/h=h /5
From above,
h /5=0.428
h =2.14 cm
Answer:
Given :-
Focal length, f = R/2 = 30/2 = 15 cm
Object Distance, u = - 20 cm
Height of object, h₁ = 5 cm
To Find :-
Image distance, v = ??
Height of image, h₂ = ??
Formula to be used :-
Mirror formula,
1/v + 1/u = 1/f
Magnification, m = h₂/h₁ = - v/u
Solution :-
Putting all the values, we get
1/v + 1/u = 1/f
⇒ 1/v + 1/- 20 = 1/15
⇒ 1/v = 1/15 + 1/20
⇒ 1/v = 4 + 3/60
⇒ 1/v = 60/7
⇒ v = 8.57 cm
Hence, the Image distance is 8.57 cm.
Now, Magnification, m = h₂/h₁ = - v/u
m = h₂/h₁ = - v/u
⇒ h₂/5 = 8.57/20
⇒ h₂ = 8.57 × 5/20
⇒ h₂ = 2.175 cm.
Hence, height of image is 2.175 cm.
The image will be virtual and erect.